Conditional on $X$,
$$
\mathbb E\!\left[\frac{ZY}{e(X)}\middle|X\right]
=\mathbb E[Y\mid Z=1,X],
\qquad
\mathbb E\!\left[\frac{(1-Z)Y}{1-e(X)}\middle|X\right]
=\mathbb E[Y\mid Z=0,X].
$$
Multiplying by $\mathbf1\{X_1=x_1\}$, taking expectations, dividing by $\mathbb P(X_1=x_1)$, and using part (i) proves the <inverse probability weighting> formula.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 221 / 4 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-221/4/a/ii/solution